Regular Singular Points and Frobenius, differential equations lesson

Near $x = 0$ the equation $x^2y'' + xy' - \tfrac49 y = 0$ , with independent variable $x$ and $y' = dy/dx$ , has the solutions $x^{2/3}$ and $x^{-2/3}$ , and the similar equation $x^2y'' + xy' = 0$ has the solution $\ln x$ . No power series in whole powers of $x$ behaves like these near $0$ , so the ordinary-point method cannot find them. Which equations still allow a series solution at a singular point, and what form must it take? This lesson classifies regular singular points, finds the leading power from the indicial equation, and shows when the second solution needs a logarithm. Throughout, the work is on the interval $x > 0$ , where $x^r = e^{r \ln x}$ and $\ln x$ are defined without any choice of branch.

Key terms

Regular singular point

Definition Let $x_0$ be a singular point of the normalized equation

y'' + p(x)\,y' + q(x)\,y = 0,

so that $p$ or $q$ is not analytic at $x_0$ . The point $x_0$ is a regular singular point if both $(x - x_0)\,p(x)$ and $(x - x_0)^2\,q(x)$ are analytic at $x_0$ , once their values at $x_0$ are defined by their limits. Otherwise $x_0$ is an irregular singular point. For a regular singular point at $x_0 = 0$ , multiplying the normalized equation by $x^2$ gives the standard form

\begin{aligned} &x^2y'' + x\,P(x)\,y' \\[0.4em] &\qquad + Q(x)\,y = 0, \end{aligned}

where $P(x) = x\,p(x)$ and $Q(x) = x^2q(x)$ are analytic at $0$ . A regular singular point at another $x_0$ moves to $0$ under the substitution $s = x - x_0$ .

In plain words At a regular singular point, $p$ has at worst a simple pole and $q$ at worst a double pole. That is the same strength of singularity as in a Cauchy-Euler equation, and it is mild enough for a modified series.

Example For $x^2y'' + xy' - \tfrac49 y = 0$ the normalized coefficients $p(x) = 1/x$ and $q(x) = -4/(9x^2)$ are not analytic at $0$ , but $x\,p(x) = 1$ and $x^2q(x) = -\tfrac49$ are constants. So $0$ is a regular singular point with $P = 1$ and $Q = -\tfrac49$ . Bessel’s equation $x^2y'' + xy' + (x^2 - \nu^2)\,y = 0$ is regular singular at $0$ as well, with $P = 1$ and $Q(x) = x^2 - \nu^2$ . By contrast, $x^3y'' + y = 0$ has $q(x) = 1/x^3$ , and $x^2q(x) = 1/x$ is not analytic at $0$ , so $0$ is an irregular singular point.

Frobenius series

Definition At a regular singular point $0$ , a Frobenius series is a trial solution on $x > 0$ of the form

\begin{aligned} y &= x^r \sum_{n=0}^{\infty} a_n x^n \\[0.4em] &= \sum_{n=0}^{\infty} a_n x^{n+r}, \end{aligned}

with $a_0 \ne 0$ , where the exponent $r$ is to be found and may be fractional or negative. The condition $a_0 \ne 0$ makes $x^r$ the actual leading power. The theorem of Frobenius says when such a solution exists. Suppose the series of $P$ and $Q$ about $0$ converge for $\lvert x \rvert < \rho$ , and the roots $r_1 \ge r_2$ of the indicial equation below are real. Then there is a solution $y_1 = x^{r_1}\sum a_n x^n$ with $a_0 = 1$ whose series converges at least for $\lvert x \rvert < \rho$ , so $y_1$ solves the equation on $0 < x < \rho$ . If $r_1 - r_2$ is not an integer, there is a second solution of the same form with exponent $r_2$ . On $x < 0$ the same statements hold with $x^r$ replaced by $\lvert x \rvert^r$ .

In plain words Let the equation choose a leading power $x^r$ first, then correct it with an ordinary power series. Close to $0$ such a solution behaves like a constant times $x^r$ .

Example For $x^2y'' + xy' - \tfrac49 y = 0$ the coefficients $P$ and $Q$ are constant, and the Frobenius series reduce to single powers. For any power,

x^2(x^r)'' + x(x^r)' = r^2x^r,

so $y = x^r$ is a solution exactly when $r^2 = \tfrac49$ , which gives $x^{2/3}$ and $x^{-2/3}$ . The singular point does not accept arbitrary initial values, so data are imposed at an ordinary point such as $x = 1$ . For $y = Ax^{2/3} + Bx^{-2/3}$ the conditions $y(1) = 3$ and $y'(1) = \tfrac23$ read $A + B = 3$ and $\tfrac23(A - B) = \tfrac23$ , so $A = 2$ , $B = 1$ and

y = 2x^{2/3} + x^{-2/3}.

Indicial equation

Definition Substitute a Frobenius series into $x^2y'' + x\,P(x)\,y' + Q(x)\,y = 0$ , where $P(x) = \sum P_k x^k$ and $Q(x) = \sum Q_k x^k$ . The lowest power $x^r$ has the coefficient $a_0F(r)$ , with

\begin{aligned} F(r) &= r(r - 1) \\[0.4em] &\quad + P(0)\,r + Q(0). \end{aligned}

Since $a_0 \ne 0$ , the exponent must satisfy the indicial equation $F(r) = 0$ , and its roots are called the indicial roots, or exponents, at the singular point. For $n \ge 1$ the coefficient of $x^{n+r}$ gives

\begin{gathered} F(r + n)a_n = -\sum_{k=0}^{n-1} c_{n,k}a_k, \\[0.4em] c_{n,k} = (k + r)P_{n-k} + Q_{n-k}. \end{gathered}

This determines $a_n$ from the earlier coefficients whenever $F(r + n) \ne 0$ . At an index where $F(r + n) = 0$ the equation for $a_n$ cannot be solved in the usual way, and that is where the difficulties with the second solution come from.

In plain words Very close to $0$ only the leading terms matter, and there the equation acts like the Cauchy-Euler equation $x^2y'' + P(0)\,xy' + Q(0)\,y = 0$ . Its exponents are the possible leading powers of the true solutions.

Example Multiplying $xy'' + y = 0$ by $x$ gives $x^2y'' + x\,y = 0$ , so $P = 0$ and $Q(x) = x$ , and $0$ is a regular singular point. The indicial equation is $r(r - 1) = 0$ , with roots $1$ and $0$ . For $r = 1$ the only nonzero term on the right is the one with $Q_1 = 1$ , and $F(n + 1) = n(n + 1)$ , so the recurrence is $a_n = -a_{n-1}/(n(n + 1))$ . With $a_0 = 1$ it gives

\begin{aligned} y_1 &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n!\,(n+1)!} \\[0.4em] &= x - \frac{x^2}{2} + \frac{x^3}{12} - \cdots, \end{aligned}

which converges for every $x$ , since $P$ and $Q$ are polynomials.

Roots differing by an integer

Definition Let the indicial roots be real with $r_1 - r_2 = N$ for a positive integer $N$ . The larger root always gives a Frobenius solution $y_1$ , because $F(r_1 + n) \ne 0$ for every $n \ge 1$ . For the smaller root the recurrence reaches $n = N$ with $F(r_2 + N) = F(r_1) = 0$ , so the equation for $a_N$ reads $0 \cdot a_N = S_N$ , where $S_N$ is the right side computed from $a_0, \dots, a_{N-1}$ . If $S_N \ne 0$ , no Frobenius solution with exponent $r_2$ exists, and the second solution contains a logarithm. If $S_N = 0$ , every value of $a_N$ works, and $r_2$ gives a second Frobenius solution with no logarithm. Changing $a_N$ only adds a multiple of $y_1$ . When $r_1 - r_2$ is not an integer, both roots give Frobenius solutions, and they are independent because their leading powers differ. An integer difference is therefore a reason to check the equation at $n = N$ , not a proof that a logarithm appears.

In plain words When the two leading powers are a whole number apart, the series that starts at the lower power eventually reaches the higher one. At that step the equation either lets it pass or forces a logarithm.

Example For $x^2y'' + xy' - \tfrac14 y = 0$ the roots are $\tfrac12$ and $-\tfrac12$ , so $N = 1$ , yet $x^{1/2}$ and $x^{-1/2}$ both solve it. Here $P$ and $Q$ are constant, so $S_1 = 0$ . For $xy'' + y = 0$ the roots $1$ and $0$ also differ by $1$ , but with $r_2 = 0$ the equation at $n = 1$ is $F(1)\,a_1 = -Q_1a_0$ , that is $0 \cdot a_1 = -a_0$ . This is impossible because $a_0 \ne 0$ , so that equation has no second Frobenius solution, and its second solution must contain $\ln x$ .

Repeated indicial root

Definition The indicial root $r_1$ is repeated when $F(r) = (r - r_1)^2$ . Then $F(r_1 + n) = n^2 \ne 0$ for every $n \ge 1$ , so the recurrence determines every coefficient from $a_0$ , and every Frobenius solution with exponent $r_1$ is a constant multiple of $y_1$ . A second independent solution therefore cannot be a Frobenius series. It has the form

y_2 = y_1 \ln x + x^{r_1} \sum_{n=1}^{\infty} b_n x^n

on $0 < x < \rho$ , so in this case the logarithm is always present.

In plain words A double root supplies only one leading power, so the second solution has to come from somewhere else. It comes from a factor $\ln x$ , just as a double root of a Cauchy-Euler equation gives $x^{r_1}$ and $x^{r_1}\ln x$ .

Example The equation $x^2y'' + xy' = 0$ has $P = 1$ and $Q = 0$ , so $F(r) = r^2$ and $0$ is a double root. The constant $y_1 = 1$ is a solution. With $v = y'$ the equation becomes $xv' + v = 0$ , so $v = C/x$ and a second solution is $y_2 = \ln x$ . Its derivatives $1/x$ and $-1/x^2$ give $x^2 \cdot (-1/x^2) + x \cdot (1/x) = 0$ . The Wronskian $W(1, \ln x) = 1 \cdot (1/x) - 0 \cdot \ln x = 1/x$ is nonzero on $x > 0$ , so $1$ and $\ln x$ form a fundamental set there. In the form above, every $b_n$ is zero.

Logarithmic second solution

Definition Let $0$ be a regular singular point with real indicial roots $r_1 \ge r_2$ , and let the series of $P$ and $Q$ converge for $\lvert x \rvert < \rho$ . If the roots are equal, a second solution on $0 < x < \rho$ has the form shown for a repeated indicial root, with the coefficient of $y_1 \ln x$ equal to $1$ . If $r_1 - r_2$ is a positive integer, a second solution on $0 < x < \rho$ is

\begin{aligned} y_2 &= C\,y_1 \ln x \\[0.4em] &\quad + x^{r_2} \sum_{n=0}^{\infty} b_n x^n \end{aligned}

with $b_0 \ne 0$ , where the constant $C$ is zero exactly when the smaller root gives a Frobenius solution. In both cases the series converge at least for $\lvert x \rvert < \rho$ , and $y_1$ and $y_2$ form a fundamental set on $0 < x < \rho$ . The coefficients are found by substituting the form into the equation. Alternatively, on an interval where $y_1$ has no zeros, reduction of order gives

y_2 = y_1 \int \frac{e^{-\int p\,dx}}{y_1^2}\,dx,

and the logarithm appears when the integrand contains a term $1/x$ . On $x < 0$ the same forms hold with $\lvert x \rvert$ in place of $x$ .

In plain words When the second leading power is missing or blocked, the equation makes up the shortfall with $\ln x$ times the first solution, plus a correcting series.

Example For $x^2y'' - xy' + y = 0$ the indicial polynomial is $r(r - 1) - r + 1 = (r - 1)^2$ , so $r = 1$ is repeated and $y_1 = x$ . Here $p(x) = -1/x$ and $e^{-\int p\,dx} = x$ , so reduction of order gives

y_2 = x \int \frac{x}{x^2}\,dx = x \ln x,

with the logarithm coming from $\int dx/x$ . For $xy'' + y = 0$ , where the smaller root $0$ is blocked, substituting the second form with $b_0 = 1$ and the free choice $b_1 = 0$ gives

\begin{aligned} y_2 &= -y_1 \ln x + 1 - \tfrac{3}{4}x^2 \\[0.4em] &\quad + \tfrac{7}{36}x^3 - \cdots, \end{aligned}

with $y_1$ from the indicial equation card. The constant is forced, since the constant term of $xy_2'' + y_2$ works out to $C + b_0$ , so $C = -1$ .

Check yourself

Classify the point $x = 0$ for $x^2y'' + 3xy' + (1 + x)\,y = 0$ and for $x^3y'' + xy' + y = 0$ . For the regular singular one, find the indicial roots and say what they predict about the second solution.

Show the answer

For the first equation $p(x) = 3/x$ and $q(x) = (1 + x)/x^2$ , so $0$ is singular, but $x\,p(x) = 3$ and $x^2q(x) = 1 + x$ are analytic, so $0$ is a regular singular point with $P = 3$ and $Q(x) = 1 + x$ . The indicial polynomial is

\begin{aligned} &r(r - 1) + 3r + 1 \\[0.4em] &\qquad = (r + 1)^2, \end{aligned}

so $-1$ is a repeated root. Up to a constant factor there is one Frobenius solution with exponent $-1$ , which works out to $y_1 = x^{-1}\sum (-1)^n x^n/(n!)^2$ , and the second solution must contain $y_1 \ln x$ . For the second equation $p(x) = 1/x^2$ , and $x\,p(x) = 1/x$ is not analytic at $0$ , so $0$ is an irregular singular point and the Frobenius theory does not apply there.

Find the indicial roots of $2x^2y'' + 3xy' - (1 + x)\,y = 0$ at $0$ . How many Frobenius solutions are guaranteed? Find the first three coefficients of each with $a_0 = 1$ .

Show the answer

Dividing by $2$ gives the standard form with $P = \tfrac32$ and $Q(x) = -\tfrac12(1 + x)$ , so

\begin{aligned} F(r) &= r(r - 1) + \tfrac32 r - \tfrac12 \\[0.4em] &= (r + 1)(r - \tfrac12). \end{aligned}

The roots $\tfrac12$ and $-1$ differ by $\tfrac32$ , which is not an integer, so there are two independent Frobenius solutions, and since $P$ and $Q$ are polynomials their series converge for all $x$ . In the original equation the coefficient of $x^{n+r}$ is $(2(n + r) - 1)(n + r + 1)\,a_n - a_{n-1}$ . For $r = \tfrac12$ this gives $a_n = a_{n-1}/(n(2n + 3))$ , so $a_1 = \tfrac15$ , $a_2 = \tfrac{1}{70}$ and

y_1 = x^{1/2}(1 + \tfrac{x}{5} + \tfrac{x^2}{70} + \cdots).

For $r = -1$ it gives $a_n = a_{n-1}/(n(2n - 3))$ , so $a_1 = -1$ , $a_2 = -\tfrac12$ and

y_2 = x^{-1}(1 - x - \tfrac{x^2}{2} - \cdots).

The indicial roots of $xy'' + 2y' + xy = 0$ at $0$ differ by an integer, and a classmate concludes that the second solution must contain $\ln x$ . Check this against the recurrence and identify both solutions.

Show the answer

Multiplying by $x$ gives $x^2y'' + 2xy' + x^2y = 0$ , so $P = 2$ and $Q(x) = x^2$ . The indicial equation $r(r - 1) + 2r = r(r + 1) = 0$ has the roots $0$ and $-1$ , with $N = 1$ . The coefficient of $x^{n+r}$ is $F(n + r)\,a_n + a_{n-2}$ , with $a_{-2} = a_{-1} = 0$ . For $r = -1$ and $n = 1$ this reads $F(0)\,a_1 = 0$ , that is $0 \cdot a_1 = 0$ , so there is no obstruction and $a_1$ is free. The classmate is wrong. With $a_1 = 0$ the recurrence $a_n = -a_{n-2}/(n(n - 1))$ makes every odd coefficient zero and gives $a_2 = -\tfrac12$ , $a_4 = \tfrac{1}{24}$ and in general $a_{2k} = (-1)^k/(2k)!$ , so

\begin{aligned} y &= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k-1}}{(2k)!} \\[0.4em] &= \frac{\cos x}{x}. \end{aligned}

The root $0$ , with $a_n = -a_{n-2}/(n(n + 1))$ and $a_1 = 0$ , gives $a_{2k} = (-1)^k/(2k + 1)!$ and the solution $\sin x / x$ . A direct check confirms both, because the equation can be written $(xy)'' + xy = 0$ , so $xy$ is a combination of $\cos x$ and $\sin x$ . Neither solution has a logarithm.

Regular Singular Points and Frobenius

Key terms

Regular singular point

Frobenius series

Indicial equation

Roots differing by an integer

Repeated indicial root

Logarithmic second solution

Check yourself

How this lesson connects

Builds on

Used later in